text:
What are the effects of an increase in altitude upon ESHP of a turboprop engine?
This is one of the questions in the ATP material:

What effect will an increase in altitude have upon the available
  equivalent shaft horsepower (ESPH) of a turboprop engine?
A) Lower air density and engine flow will cause a decrease in power
  B) Higher propeller efficiency will cause an increase in usable
  power (ESHP) and thrust
  C) Power will remain the same but propeller
  efficiency will decrease

I chose (C) but the correct answer was (A). Could anyone explain why? Isn't the whole purpose of a turbocharged engine to sustain engine power at high altitude (below critical altitude)? Why does lower density at high altitude cause a decrease in power?

You have confused a turboprop engine with a turbocharged engine. They are very different and have very little in common.
A turboprop engine is powered by a turbine engine geared to a propeller. Turbine engines have less power available at higher altitudes so answer "A" is correct.
A turbocharged engine is a piston engine which is fitted with a small compressor driven by exhaust gas pressure. The compressor (turbocharger) can increase manifold pressure to restore sea level power at altitude. (Turbo-normalized) A turbo charger can also supply greater than sea level manifold pressure for more power. (turbo-supercharged)
Both types of turbocharged piston engines can maintain full rated power as they climb, but they will eventually reach a limit where full power can no longer be maintained.

Okay I might have confused the turboprop with turbocharger, but doesn't the turboprop also have a compressor in the engine system? If so, can't it maintain air pressure flowing into the turbine at altitude?

The compressor compresses with a certain ratio, let's say 9:1. Air exits the compressor at 9 times the pressure that it entered in, at higher altitude the entry pressure is lower and so is the exit pressure.

@Koyovis When you say the compressor in your comment, is it limited to the compressor of turboprops or does it include that of turbochargers?

I was referring to the turbine compressor. The turbo charger has an arrangement where a turbine in the gas exhaust drives an intake air compressor, very similar to a turbine compressor arrangement actually. A ‘turbo-normalized’ turbocharger maintains sea level horsepower performance as it reaches higher altitudes.

@lemonincider, actually, turboprop and turbocharged engines are very similar and have almost everything in common. Both (and naturally-aspirated ones too) are limited by air density and their power decreases with altitude. However, all kinds of engines are also limited by other factors—maximum pressure and temperature the engine can withstand—when the density is above certain value, so below corresponding altitude, they have constant power (are “flat rated”). The only difference is the typical altitude to which different kinds are flat rated.

@MikeSowsun, no, compressor of a turbine engine does not lose efficiency as altitude increases. It still provides the same pressure ratio, just like a piston compressor (in a piston engine) does. However, as it is a ratio, half the density in intake still means half the working gas mass in the combustion chamber and that limits the power. Losing efficiency would mean that it would need to extract more energy from the exhaust (higher pressure ratio across the turbine) to add the same energy to the intake (same ratio across the compressor), but that is not the case.

Because the performance of a turboprop engine, like any other air cycle heat engine,  is directly proportional to the density of the air moving through the engine.   Denser air means more working fluid in a greater amount of energy can be obtained from it by combusting it with fuel.   As you rise in altitude, the air becomes less dense and the amount of work that can be obtain from a given volumetric flow rate of air becomes less and less the higher you climb.   This also lowers the equivalent shaft horse power that one can obtain from the turboprop engine, as it both reduces the shaft horse power to the propeller as well as the jadditional et thrust from the exhaust.
Manufacturers of turboprop engine sometimes compensate for this  by flat rating the turboprop engine. That is, the fuel control unit for the turboprop presets the amount of output power that the engine will produce.   The gas core of the turboprop may be able to produce much more power at sea level or at lower altitudes than you can obtain with a full throttle command from the cockpit.  The maximum amount of power the gas core on a flat rated turboprop can produce at sea level at standard atmosphere is referred to as  the engine's thermodynamic power rating.    A flat rated engine will continue to produce a uniform power output as the aircraft continues to climb until the flat rated power output is equal to the thermodynamic power output of the engine at the current flight level.   If the aircraft continues  to climb, the engine's power output will begin to decrease.

The turboprop engine produces less power at altitude, because the mass stream through the engine is smaller: there is less air. The turbocharged piston engine also produces less power at altitude due to less air entering the cylinder. Both have a compressor, and some effects of altitude on power generation are indeed similar, but the design conditions differ due to the great differentiator: weight.
A turboprop engine's weight increases much less as a function of max power at sea level, than that of a piston engine. The turboprop can be designed for cruise conditions, and take the surplus of power at sea level as either:

a bonus, so that much more power is available at sea level and power output decreases gradually with altitude;
a liability for the gearbox, so that power limitations at lower altitude are applied by the FADEC.

If the same approach was followed for a piston engine, the weight penalty would be much higher than for the turbine engine. So piston engines are designed for maximum power at sea level, and this power is maintained as long as possible with increasing altitude, by pumping in more air through the super/turbocharger.
Note that the turboprop with derated power and the turbocharged piston have very similar characteristics, both deliver a constant power as altitude increases, up to the critical air pressure...
The weight advantage of turboprops over piston engines was demonstrated in the 1960s by the conversion of the Cessna Skymaster into the Conroy Stolifter: the two Skymaster piston engines produced 155 hp less and weighed 117 kg more than the single TPE-311 turbine engine that replaced them.
 Image source
 Image source


text:
How does downwash change relative wind?
I have a question regarding the production mechanism of induced drag.
I understand the downwash aft of the trailing edge (steaming from wingtip vortices) causes the induced drag by changing the relative flow (blue line) into the effective relative airflow (red line), and thus by bending the lift vector rearward (L to Leff).

My question here is: How does the downwash, that has already passed behind the trailing edge of the airfoil, influence the airflow upstream, causing the entire relative airflow to bend upward as in the picture.
Or is it just an imaginary concept to explain the induced drag?

See Why does the vortex created by wing affects its own angle of attack?

"The influence of the downwash actually extends ahead of the wing. The air ahead begins to be pulled down in response to the flow behind the wing, in proportion to the downwash velocity, even before the wing arrives". (Source.) The why is viscosity, and the initiator is the downwash of the tip vortices.

@mins: Your source is incorrect: Ahead of the wing the flow moves up. This can be explained by vorticity (the bound vortex adds an up component there) and by physics: The suction on top of the airfoil sucks in more air. See this answer for a better explanation. I recommend to doubt everything you read on the Interwebs (including my answers).

@PeterKämpf: With due respect, I'm not convinced because it seems to me we are mixing images of 2D sections of airfoil without tip vortices downwash effect and 3D figures with it. So let's put the two together. In a) the 2D section with clearly a up-down path like in your picture. In b) 3D actual flow with the inner downwash of the tip vortices: I see clearly a descending pattern. Image from Aircraft Flight (Barnard, Philpott)

The above drawing just appeared on the Quora Site.  It fails. There is a tendency to complicate this and use misconceptions.  If you are willing to accept that the yop-bottom pressure difference causes the tip-spill thus contributing to (not fully causing) the tip vortex, THEN, you should also be able to understand that these same pressures cause the up-wash ahead of the wing.

Following up on Bernoulli's work in the mid 1700s, it was Euler who determined that a Pressure Gradient causes acceleration of fluid from the higher toward the (relatively)  lower pressure areas.  There is no indication in Bernoulli's notes that he understood any cause and effect relationships (John Anderson).

If the under wing pressure is higher than the above wing pressure, then ahead of the leading edge there WILL be upward acceleration and THIS IS OBSERVED in real airflows; NOT amateur drawings! In addition , the above drawing fails because it ignores the up-wash.   The up-wash has the effect of an increase in AoA.

This is evidenced by the sharper curve in the upper airflow around and above the leading edge, because of this up-wash upward component.  Look only at real flows, such as Babinsky's Cambridge video.   That above wing sharper curved flow is evidence that the pressure change (drop) is greater than below the wing (increase) and, THEREFORE,  a larger contributor to the top-bottom lift causing pressure difference. Try this on for size: Understanding lift Correctly: https://rxesywwbdscllwpn.quora.com/

How does the downwash, that has already passed behind the trailing edge of the airfoil, influence the airflow upstream, causing the entire relative airflow to bend upward as in the picture.

The downwash is behind the trailing edge, but the acceleration of the air downwards happens above and below the wing. The highest acceleration is around the quarter-chord where the centre of pressure is.
In subsonic flow, the pressure changes do propagate upstream due to how the various flow equations work. The sharp trailing edge is indeed an important feature of efficient wings that affects how the flow behaves around the wing.

Or is it just an imaginary concept to explain the induced drag?

It isn't. But there are many ways to explain it. I prefer using the conservation of momentum and energy (that I explained here).
Note that for supersonic wing, the mechanism changes to wave drag. The conservation argument still holds and explains why there can't be lift without drag, but now that the pressure changes do not propagate upstream the pressure field looks differently.

The downwash doesn't cause an upward bend.  The higher pressure under the wing compared to above it is the source of an upward acceleration both ahead of the wing AND around the tips.  A Pressure Gradient is the Cause of an acceleration - from low pressure to relatively higher pressure.  Pressure Gradient causes fluid acceleration.(Euler).

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text:
61 vs 141: What's better for future US ATP?
The pros and cons of each are all over the internet. Price and flexibility, which of the two routes looks better when applying for an airline regional or legacy, does it matter? More importantly which better prepares you?

There is no concrete answer to your question, only opinion. As a person who has hired pilots into single engine piston to twin turboprops (charter world): absolutely nobody cares. Take the most affordable route that gets you to revenue in the least amount of time. I found this to be pt61 because it let me build aggressive timelines for training. 141 was too regimented and the ground school sessions were just something I would do anyway on my own for free. If you need structure then 141 may be better. If you want to design your own timeline then 61 may be better.

Also, the minimum times quoted by virtually every flight school are waaaay too optimistic for a typical training schedule and student.

One big difference is that going to a flight school as part of a two or four year degree at many approved colleges can reduce the flight time you need before getting an ATP.

I do not agree the question should be closed as "primarily opinion based". To the OP, the difference between part 61 and part 141 is unclear, hence the question. If both choices are equally acceptable, then the answer to the question should be "it does not matter".

Regarding the degree, go for a non-aviation major.  Knock out your ratings through your CFI and instruct your junior and senior years.  Easily doable if you have the money or loans.

@acpilot Some companies do like to see an aviation related degree. And I understand that to be eligible for the R-ATP (with lower time required) the 2/4 yr degree typically needs to be an aviation degree.

@JonathanWalters.  I have never seen an aviation employer worth working for care at all about a college major.  That a pilot has a 4yr degree is the only box that is checked.  If a career change is the motivation for an ATP then, sure, go 141.  If the prospective ATP is an 18yo looking for a career path they should not get an aviation degree.  The pt61 route and a practical degree can get the student affordable training, a CFI job, the flt time for an ATP as a CFI and a real-world fallback plan all in one 4yr span.

Great, yes my first choice is a Part 61 due to affordability and flexibility/length of training. It's also way closer to me. @acpilot Thanks.

Are you doing a career change or starting college?

Either one is fine, though I'd make sure you seek a good instructor for a part 61 and follow a part 141 syllabus (available through ASA, Jeppesen, King Schools, etc.) when you do yours to verify you covered all the bases.
At that stage of your training, future employers really won't care that much whether you did your training as a 61 or 141 student.  They are going to be much more concerned with your professional flying record and later certification and checkrides.

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text:
What is implied about altitude and the VDP by this Visual Flight Segment arrow in the profile view example in the Aeronautical Chart User's Guide?
This is out of the Aeronautical Chart User's Guide.

The arrow for the visual segment is depicted as being level flight.  Is that what is intended to be conveyed?  If so, how would you know what altitude?  Would it just be the DA?
LNAV/VNAV would only be second to LPV in vertically-guided precision.  Does this imply that the VDP applies only to an LPV minima, or does the VDP apply to both LNAV/VNAV and LPV minima regardless of the visual segment arrow for LNAV/VNAV??

This is extracted from the RNAV (GPS) RWY 23 of WESTERN NEBRASKA RGNL/WILLIAM B HEILIG FIELD (BFF) in Scottsbluff, Nebraska.

A VDP (Visual Descent Point) is only used in conjunction with an MDA (Minimum Descent Altitude). Do not mistake this for a DA (Decision Altitude). If visual contact with the runway environment is not made once a pilot reaches MDA, the pilot must continue at the same altitude on the approach course until reaching the MAP (Missed Approach Point). The VDP is the point along the course where a normal descent has been calculated as possible using normal procedures and maneuvers. It is a recommended point where the pilot should make the decision to go missed once reaching the MAP. That means that even though the pilot has decided to go missed, they will continue to fly past the VDP, straight and level, until the MAP. The key word here is “recommended”.
At a DA, if the pilot has not made visual contact with the ground, the pilot is required to decide to go missed. In most cases, that means the pilot would initiate the missed approach procedure immediately before reaching the DA. Since there are two DAs for the LNAV/VNAV, depending on if the local altimeter setting is available, there are two distances to the MAP. This is conditional upon the pilot maintaining glideslope all the way from the FAF (Final Approach Fix).
In this case, upon making visual contact with the ground, the pilot can choose to make a Visual Approach. There is no altitude requirement unless specifically charted. Cloud clearance and visual obstacle clearance must be maintained.
According to the users guide, it appears a pilot on the approach can fly visually past the point of decision if they have broken free from the clouds.
The Aeronautical Chart Users Guide states...

Visual Flight Path
  Instrument approach procedures, including Copter approach procedures, that terminate or have missed approaches prior to the airport, and are authorized to proceed visual, shall depict the visual segment by the dashed line symbol from the missed approach point to the airport. The note “Fly visual” (“Proceed visually” on Copter procedures) along with the bearing and distance shall be shown leadered to the visual flight path.
  RNAV charts sometimes have visual flight for LNAV/VNAV minima which do not start at the missed approach point. An additional note indicating “LNAV/VNAV” will be placed above the note.

Furthermore the AIM 5-4-5-l states...

l. In isolated cases, an IAP may contain a published visual flight path. These procedures are annotated “Fly Visual to Airport” or “Fly Visual.” A dashed arrow indicating the visual flight path will be included in the profile and plan views with an approximate heading and distance to the end of the runway.

The depicted ground track associated with the “Fly Visual to Airport” segment should be flown as a “Dead Reckoning” course. When executing the “Fly Visual to Airport” segment, the flight visibility must not be less than that prescribed in the IAP; the pilot must remain clear of clouds and proceed to the airport maintaining visual contact with the ground. Altitude on the visual flight path is at the discretion of the pilot, and it is the responsibility of the pilot to visually acquire and avoid obstacles in the “Fly Visual to Airport” segment.
Missed approach obstacle clearance is assured only if the missed approach is commenced at the published MAP. Before initiating an IAP that contains a “Fly Visual to Airport” segment, the pilot should have preplanned climb out options based on aircraft performance and terrain features. Obstacle clearance is the responsibility of the pilot when the approach is continued beyond the MAP.

Note: The FAA Administrator retains the authority to approve instrument approach procedures where the pilot may not necessarily have one of the visual references specified in 14 CFR §91.175 and related rules. It is not a function of procedure design to ensure compliance with §91.175. The annotation “Fly Visual to Airport” provides relief from §91.175 requirements that the pilot have distinctly visible and identifiable visual references prior to descent below MDA/DA.

My reasoning for saying the VDP is recommended and not required is that the VDP is found in the AIM.

AIM 5-4-5
  h. The Visual Descent Point (VDP), identified by the symbol (V), is a defined point on the final approach course of a nonprecision straight-in approach procedure from which a stabilized visual descent from the MDA to the runway touchdown point may be commenced. The pilot should not descend below the MDA prior to reaching the VDP. The VDP will be identified by DME or RNAV along-track distance to the MAP. The VDP distance is based on the lowest MDA published on the IAP and harmonized with the angle of the visual glide slope indicator (VGSI) (if installed) or the procedure VDA (if no VGSI is installed). A VDP may not be published under certain circumstances which may result in a destabilized descent between the MDA and the runway touchdown point. Such circumstances include an obstacle penetrating the visual surface between the MDA and runway threshold, lack of distance measuring capability, or the procedure design prevents a VDP to be identified.

VGSI systems may be used as a visual aid to the pilot to determine if the aircraft is in a position to make a stabilized descent from the MDA. When the visibility is close to minimums, the VGSI may not be visible at the VDP due to its location beyond the MAP.
Pilots not equipped to receive the VDP should fly the approach procedure as though no VDP had been provided.
On a straight-in nonprecision IAP, descent below the MDA between the VDP and the MAP may be inadvisable or impossible. Aircraft speed, height above the runway, descent rate, amount of turn, and runway length are some of the factors which must be considered by the pilot to determine if a safe descent and landing can be accomplished.

That's right.  Sorry dumb question.  What I'm really trying to get at here is more about the visual segment arrow, not the VDP.  Could you edit the answer to address that as well?

Edits are in bold.

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text:
These "FICON" NOTAM's contractions aren't in the FAA contraction list. What do they mean?
These NOTAMs for KSGF state "Snow" when the icon on the left is hovered over with a mouse.  The contraction "FICON" does  not appear in the Notam Contractions list.
It is currently the middle of June in nearby Oklahoma, and it is hot here, so a NOTAM related to snow doesn't make any sense.
What are these NOTAMs telling us?

FAA JO 7930.107 FICON Reporting

@RonBeyer a few appropriate excerpts from that PDF and you've got yourself an answer...

KSGF is in Missouri. Not that it makes any difference to your question.

I've wondered that before, why FICON does not appear in the NOTAMs contractions list.

@TomMcW thanks, I wasn't paying attention.

FICON is outlined in FAA JO 7930.107 "FICON REPORTING".
FICON is the "FIeld CONditions reporting. It identifies information that used to be called "braking action", but is now a little more succinct.
The conditions for the runway are reported in thirds, [First Third]/[Middle Third]/[Last Third], with the following data outlined on page 6:

So 5/5/5 means that the runway is covered with frost or is wet and is "damp or 1/8" or less of standing water". Pilots used to report this as "good" braking action.


text:
How does CDI scaling transition from terminal to approach sensitivity?
I read in the AIM that WAAS vector modes will provide a linear course width of +/-1NM prior to the FAF and then provide a fixed width of +/-0.3NM to the point where an ILS splay would be 0.3NM and then tapers down to 700 ft total width at the runway threshold.
How smooth is the transition from the 1NM segment to the +/-0.3NM segment? Is it time-based, distance-based, or something else? Is it smooth at all, or is it actually an abrupt change?
Are all manufacturers following a common standard on how these scale transitions work?

I'd like to note that my original missed point was that the transition was "Smooth" over that 2NM period as the question cites the same point in the AIM that I've used as an answer later on with my own improved overall understanding.

Beginning 2NM prior to reaching the final approach fix of an activated approach, the GPS will smoothly taper down the CDI scale from 1-NM full-scale deflection (from the center) to 0.3-NM at full-deflection over that 2-NM flight distance and reach a scale of 0.3-NM as the final approach fix is reached.
AIM Page 1-1-25

"When within 2 NM of the Final Approach Waypoint (FAWP) with the
approach mode armed, the approach mode will switch to active, which
results in RAIM and CDI changing to approach sensitivity. Beginning 2
NM prior to the FAWP, the full scale CDI sensitivity will smoothly
change from ±1 NM to ±0.3 NM at the FAWP. As sensitivity changes from
±1 NM to ±0.3 NM approaching the FAWP, with the CDI not centered, the
corresponding increase in CDI displacement may give the impression
that the aircraft is moving further away from the intended course even
though it is on an acceptable intercept heading. Referencing the
digital track displacement information (cross track error), if it is
available in the approach mode, may help the pilot remain position
oriented in this situation. Being established on the final approach
course prior to the beginning of the sensitivity change at 2 NM will
help prevent problems in interpreting the CDI display during ramp
down. Therefore, requesting or accepting vectors which will cause the
aircraft to intercept the final approach course within 2 NM of the
FAWP is not recommended."


text:
Where does the difference between calculated and actual groundspeed come from?
I'm a paragliding pilot. At school we learn that the trim speed of a paraglider is about 38km/h airspeed, the accelerated airspeed is about 45km/h and the minimum airspeed is 22km/h - being inside the weight range of the wing.
If I fly with a headwind of 25km/h, my groundspeed should be: 38-25=13km/h.
Well, from my flying experience (1200 flights in total) with that kind of headwind I never went forward +13km/h groundspeed but much less, near the minimum relative speed (22km/h).
Why there's a difference between the calculated groundspeed 13km/h and the actual groundspeed?

I second @ymb1's question; I don't see how you could know what your actual headwind component was at any given location without having at least an airspeed indicator which allowed you to compare computed true airspeed to GPS groundspeed. When I have flown with a strong headwind resulting in near zero groundspeed, I have always found that the headwind approximated my true airspeed.

@ymb1: At the landing area (beach) the Vw was about 20km/h calculated with a weather station.

@Paragliding: The wind is generally stronger at altitude than at ground level.

I have a little paragliding experience. While only a tiny fraction of yours, I think I can answer your question, which boils down to: data accuracy.
The trim speed of a fixed-wing aircraft would depend on its weight. Without accurately measuring the pilot's weight (FYI the measurement in small aircrafts is precise to pounds / kilograms) and plotting that against a chart, the speed you were told in school is only an approximation.
I (like others in the comments also) question the accuracy of your speed data.

To accurately measure forward airspeed, you'd need a pitot tube pointing in the right direction.
Or you can use a wind measuring device which measures speed and direction
To measure your ground speed accurately you'd need a GPS.
To combine the two (wind speed + ground speed), you'd need some sort of computer to do the computation, either in real time or with recorded data.
The ground station only measure wind speed at ground level. When you're up in the air (albeit relatively low compared to aircraft), the wind can be different.

For sure data accuracy is one of the problems and the speed component at ground level. But in my experience, flying forward with 25km/h or more into the wind, makes it really harder to have some Vg. Thus, or the paragliding data from the manufacturers are not real or the formula Vg=Va-Vw has some limitations due to many other components. Amon others, again in school they teach that 25km/h headwind is the limit for us to fly forward.

@Paragliding the only situation when the formula is wrong is when the speed concerned are comparable to the speed of light. Aircraft manufacturers use detailed lab tests (or computer simulations nowadays) to come up with a detailed chart where you can just plug in the weight on one axis and you'll get the exact speed on the other. Therefore the data you're getting is inaccurate is the most likely explanation.

It's inaccurate because wind measured at ground level is usually much different as you get away from the surface.

It depends on the difference between the predicted wind speed for the winds aloft vs what was actually up at that altitude at a given time.
It also depends on doing your winds aloft calculations correctly and accounting for both the wind speed AND angle relative to the direction of flight.  Take the winds at a 60° angle, our headwind component is only going to be half the total wind speed, etc.
Then there are installation and calibration errors.  I don't know what a paraglider has on it in order to sense airspeed but a primitive pitot static system of some sort would be a reasonable guess.  I don't know what kind of error may be associated with that as well.

I know the value of my trim speed from the paraglider manufacture 37-39km/h. Then I fly into the wind, no brakes applied, no position change, 50m above the landing area in which the Vw is 18-20km/h. Then somebody down calculates my shadow path from point A to point B, 100m straight line. Based on theory, I should have a Vg of 38-20=18km/h. Based on what we measured I had a Vg of 8-10km/h. The difference is quite important for a paraglider. We did some other tests in windy conditions and the result is that the Vg is less then Vg=Va-Vw. Even in calm conditions the Vg is never equal to Va.

The wind speed, and direction, can be significantly different on the ground versus 50 meters in the air. When the air is close to the ground it is subject to ground friction. Wind is usually stronger as you go higher which would correlate with your measurements.

You are right but even flying in nil wind very early in the morning the Vg is not equal to Va calculated with the same method. I believe that one of the reasons has to be found on the fact that we do not have an engine to adjust thrust (just our body) and in real life the wind force/typology is never ideal, laminar and uniform.

No wind on the ground doesn't mean none at altitude. I remember talking off on a calm day and when I transitioned to slow flight at 2000', I was actually flying backwards relative to the ground, which meant I had a 45kt headwind.

Actual groundspeed is not in itself a useful (or deterministic) number while flying. It's useful as a concept.
You get your measured ground speed from your instruments / sensors. They'll give you a reading that's as accurate as you can get. This isn't enough for most flights however. Even in 2017, our data is still faulty.
In engineering, the difference between actual and measured anything is called the error. In this case, groundspeed has an error while using instruments. Errors usually come in ranges. For example, in paragliding, you may see a measured groundspeed of 12mph, with an error of +/-1, meaning that the actual ground speed can be between 11 and 13 mph. Error can also have two parameters as it's range. For example, you may be going along at 12 mph, and have an error of +2/-0, meaning that actual speed is between 12 and 14 mph. usually one of the parameters in a two-parameter error will be zero. If they're both non-zero, non-equal values, you're going into probability distributions, and other mathematics of the sort.


text:
Can you identify this helicopter?
I saw this helicopter in Hungary. It might be an Airbus H225M, but I'm not sure. Can you confirm or identify it correctly?

Unknown helicopter seen in Hungary
What is the red spot on the side, under the tail boom?

Wow, you went through the trouble of panning along with one of the rotor blades, so it would be sharp whilst the helicopter is motion-blurred! Interesting as this demonstrates that the tail rotor turns counterclockwise/backwards.

This is the Airbus Helicopters H225 (previously Eurocopter EC225 Super Puma). What most notably differentiates this version from the earlier H215, are the rotor assemblies: H225 has a five blade main rotor and four blade tail rotor, H215 the blade counts are "opposite", being four for the main rotor and five for the tail rotor. Otherwise, the models are externally quite similar.
As for the red object on the side: thanks for Affe for pointing me in the right direction, the device is an Emergency Helicopter Locator. Below is a picture taken from Heli-One mods catalogue 2018, page 67, which shows the device installed on Sikorsky S-76 A/C.

The brochure describes the device as follows:

The system features
a locator beacon containing the transmitter and antenna. This is automatically
activated and released in the event of a crash,
or activation of the water sensor, or manual deployment by the flight crew. The
beacon transmits on all international distress frequencies and is compatible with the
COSPAS/SARSAT system. Its GPS input will allow the beacon to transmit the last
known position of the aircraft in distress, saving lives.

jpe61 thanks for yor answer, but what is red point on side?

Let me verify my thoughts, I'll get back at ya.

This is a tricky one, on H225 it's on the side of a fuel tank, but I'm unable to verify that the doohickey is (or is not) fuel related. Some H225's have it, some don't. It's not specifically civilian or military stuff either.

It's a cpi/emergency locator beacon.

That’s an Airbus Helicopters H225 Super Puma.

I think that should be either Airbus Helicopters H215, or the old designation Eurocopter AS332 Super Puma.

Good catch! Fixed it.


text:
Is the compressor required on jet engines? Can air be rammed into the turbine?
In the context of a common airliner (say Boeing 777), when the airplane is on the ground, and the turbines are on, air is getting sucked in and compressed/bypassed. The engine has to do work to accelerate steady air in front of the engine in order to get enough in per second.
But I can imagine a speed at which a flying airplane no longer has to do work to accelerate air into the engine, as enough is getting rammed in per second.
My questions are:

Do any turbofan powered airplanes go fast enough that they no longer need to "suck" air in to maintain power?

At what point, if any, does air stop getting sucked into the turbine and is just getting rammed in?

Some high-supersonic jets work like this. But for almost all jet engines the air velocity at the entrance of the engine and the air velocity that the aircraft sees is different. So there is always some acceleration of the airflow.

Congratulations! You have reinvented the ramjet.

@StephenS All we need now is a Proton Energy Pill.

Your questions

At what point (if any) does air stop getting sucked into the turbine and is just getting rammed in?

On a CFM56-7B engine (Boeing 737 NG) pressure is provided by the fan, the low pressure compressor and the high pressure compressor with respective pressure ratios of about 1, 3 and 9, that is pressure is increased by a factor of 28 before air is delivered to the combustor for a cruise speed of about Mach 0.8.
This required compression ratio is indeed reduced as airspeed increases and pressure from speed increases. This natural compression occurring at the engine inlet is called ram pressure:

Ram compression ratio vs airspeed, source
About Mach 2.5 or 3, air can be delivered in sufficient quantity to the combustor without a compressor. The fan is not needed anymore, all air entering the engine is mixed with fuel. The turbine can be removed too as there is nothing to spin anymore.
Such engines without compressors are named ramjets and scramjets. They have no rotors or stators. In theory a ramjet could work starting at Mach 0.5, but it doesn't produce any useful thrust:

Rocket, turbojet, ramjet, scramjet efficiency, source

Do any turbofan powered airplanes go fast enough that they no longer need to "suck" air in to maintain power?

A turbofan combines a core engine and a fan. The fan produces most of the thrust, the core engine mostly turns the fan.
Fans cannot work at supersonic speed, they wouldn't be efficient. therefore when the compressor is not required, airspeed is too fast for a fan. Actually neither the ramjet nor the scramjet have a fan, they are pure reaction engines.

Ramjet
The first engines to operate on this principle is the ramjet. The difficulty is to maintain a combustion at this speed (this is a 3,000 km/h wind), to solve this problem, a ramjet slows air down before it is mixed with fuel in the combustor.

Ramjet princple, Wikipedia

Nord 1500 Griffon II, one of the first ramjet aircraft, source
Scramjet
With some improvements, supersonic burners have been built, though very inefficient ones. This engine is known as scramjet. It works at higher speeds than a ramjet.

Scramjet principle, Wikipedia
Need a little help
However no ramjet, and even less a scramjet, can produce static thrust. Both need some initial forward speed for to start the combustion. They can be launched from another aircraft:

Leduc 010 on top of SE-161 for an air-launch, source
Aircraft can be fitted with an additional subsonic engine, a convertible engine (turbojet or ramjet) or with rocket boosters.

J-58, convertible turbojet/ramjet used on SR-71 and other aircraft, source
In spite of their inconveniences, ramjets and scramjets are used on a few fast military aircraft, but their kingdom is missiles where larger airspeed often means larger altitude and extended range.

"Actually supersonic (and hypersonic) engines have no bypass." This seems to imply that all supersonic engines are turbojets (and not turbofans).  But for example F-16 fighter has a max speed of around Mach 2, it is powered by F110-129 which does have bypass (now it is low bypass, only 0.76:1, but still it is bypass).  Maybe you just meant to say hypersonic here?

@DanielK, I made a mistake in the wording, what I wanted to say (I'll update) is neither the ramjet (subsonic combustion) nor the scramjet (hypersonic combusion) have a fan.

What's the X-15 photo doing here?  That was a pure rocket spaceplane, never operated with any kind of air-breathing jet.  An SR-71 would be a better illustration, with its turbo-ramjet (switching over between turbojet and ramjet somewhere above Mach 1).

Don't forget hybrid engines like the turbo-ramjet

Note that ramjets has been applied in the past (late fifties) to powering helicopters by propelling the tips of the rotor blades. But the problem was the noise... and fuel consumption. Awesome sight for spectators on the ground of night flights!

@0scar: See also.

@mins Thanks, I wasn't aware of a predecessor!

The link underneath the lsp efficiency plot seems to not work.

@Koyovis. Thanks, fixed. Unfortunately a direct link to the pdf content.

Like any kind of vacuum system, vacuums cannot “suck“ anything. They just remove air from an enclosed space. This causes a pressure differential between the evacuated space and the surroundings.  The higher air pressure outside forces air into the cassion; it is not “sucked” in.
This is similar to the diffuser on a jet engine. The fan or compressor is drawing in air which creates a low pressure region in the diffuser, thereby causing air at local atmospheric to rush into the space.
What you’re describing here would be the concept of a ramjet, where the stagnation pressure of the diffuser is enough to compress the incoming air to the point where it can be burned and produce useful thrust without the need for augmentation by a mechanical compressor.  Turbofans and turbojets cannot do this efficiently at their operating speed ranges but do get a performance boost via diffuser compression of incoming air.

It is certainly possible to 'ram' air into a turbine. The maximum pressure that can be achieved is called the 'stagnation pressure', which is the point at which all incoming kinetic energy is converted into pressure energy.
For some applications, this is good enough. For example, the Meredith effect is responsible for a slight amount of net thrust in the radiator of the P-51 Mustang. For other applications, like a (subsonic) turbofan, while it helps a little bit in increasing the compression ratio, it cannot replace the compression section. The reason is that the downstream pressure will accelerate air in front of the engine away and outwards: air will simply 'spill' past the engine. Engines can be designed such that they exactly neither 'suck' nor 'spill' air at the design point.
The good news is that in supersonic applications, the downstream pressure cannot 'travel' upstream (because pressure travels upstream at the speed of sound). Here, incoming air is truly 'rammed' and has no choice but to undergo compression. (Sc)ramjets (and in lesser extent, inlets of supersonic turbojets) are based on this principle, but to achieve a meaningful compression ratio, they ideally would travel at hypersonic speeds for which they need a subsonic-capable engine.


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Is induced drag essentially nothing more than a specific type of form drag?
If I were standing outside on a windy day holding a flat rectangular object such as a food tray up in the wind, orthogonal to the direction of flow, the drag created would be classified, mostly, as form drag; a differential in pressure from the higher pressure upwind side of the tray and its lower pressure downwind side.
If an airplane's wing began at a 0 degree angle of attack and were slowly rotated to a 90 degree angle of attack, lift would increase until the critical angle of attack is exceeded, but the coefficient of drag would continue to increase up to 90 degrees.
I assume the drag created by the wing at 90 degrees is much like the form drag created by the tray in the wind.
Is induced drag nothing more than form drag with a different classification due to its usual nature of discussion in regards to lift?

No, by definition it isn't.
There is drag. Drag is caused by different physical phenomena. According to the cause, it is classified to:
Induced drag
Induced drag is side-effect of generating lift over finite wing span.
Lift is an upward force that the air exerts on the wing. By principle of action and reaction (Newton's third law of motion), the wing exerts downward force of equal magnitude on the air, and since the air is free to move, that force accelerates the air downwards.
Now there are two equivalent arguments why this causes drag:

In the frame of reference of the wing, the wing is not doing any work (because it is not moving), so it can't change kinetic energy of the air, and therefore can't change its total velocity. Since the vertical component of that velocity increases, the horizontal component has to decrease and this requires forward force from the wing. The backward reaction to that force is the induced drag.
In the frame of reference of the oncoming air, the kinetic energy does increase. Therefore the wing has to do some work on the air, which means applying force in direction of its motion. The backward reaction to that force is the induced drag.

The explanations are both equally true, just using different frame of reference.
Induced drag decreases with (square of) speed (for constant lift), because at higher speed there is more air to accelerate, so it only needs to be accelerated by less.
Induced drag is also independent of cross-section. If you cancel lift by moving elevator or aileron up, the cross-section may not change at all (the flap might still be behind the wing), but the induced drag will vanish.
By the way, it follows from this that induced drag of wing at 90 degrees is zero, because it is not producing any lift. The form drag is, of course, huge.
Form drag
Form drag is caused by imperfect pressure recovery behind the body. At the front of the object moving through fluid the fluid slows down, which comes with increase of pressure. In ideal laminar flow the fluid closes behind the object with the same slow speed and same high pressure. But as the object moves faster, the inertia causes the flow to separate and form a turbulent area behind the object where the pressure remains low. This pressure difference cause the form drag.
The form drag depends on cross-section and shape of the object. It always increases with speed.
Interference drag
When complex structure, like an aircraft, is analysed, the form drag of each component (wing, fuselage, empennage etc.) is first analysed independently. But as the components are assembled together, the pressure fields around the parts negatively affect each other and this increase in drag is then called interference drag.
Skin drag
Skin drag is caused by the friction between the surface of the object and the fluid. The skin drag depends mainly on wetted surface of the object. It increases with speed in similar way to the form drag, so for practical purposes it is usually combined with the form drag into one term, the parasitic drag.
Wave drag
As the object moving through fluid approaches the speed of sound, the flow exceeds it as it tries to avoid the object and forms a shock wave. Behind the shock wave the pressure is much higher and this increase is not balanced by any increase on the aft part of the object, therefore causing large amount of drag.
This kind of drag only appears as the flow speed approaches speed of sound, but then it quickly grows larger than the other forms.

Induced drag is different as it decreases with speed, but so does the cross-section of the object striking the air and creating the drag. I wonder: if a specific angle of attack were forced to remain constant upon a wing in a controlled setting, would the drag not increase with speed?  If it would, that would also lead me to think it is just a type of form drag.

@RyanMortensen, if you kept the same angle of attack and increased speed, the lift would increase and the induced drag would increase with it. But e.g. by changing camber, you can keep the cross-section the same while changing induced drag significantly. But ultimately, the argument is that the definition is different. I have rewritten the answer to include more detailed definition of the other forms.

Your paragraph on interference drag is absolutely the best, clear, concise explanation I have ever read for it.  I really like your answer the way it is written now, but I will give some time for others before concluding the Q&A process.

Your two bullet points are really confusing me. The way I've come to understand it is that the vector of the resultant force (the complete aerodynamic force on the wing due to AoA) is not perpendicular to the airflow but angled slightly backward because the air is being forced downward and forward. If you break the vector angle down into vertical and horizontal components (with respect to the airflow) the vertical is lift and the horizontal is induced drag. We might be saying the same thing, just different ways, but I'm not grasping your explanation.

@TomMcW, well, the force is angled, but you need to know why it is angled. There might be some direct argument from the lift theory, but that is rather complicated. However, simple argument can be made from just laws of motion and conservation of energy and that is what I am trying to do here (conservation of energy is a funny physical law; it holds in all reference frames, but the energies look very different in each, so we have different, equally valid, descriptions depending on which reference frame we use).

You state: The backward reaction to that force is the induced drag but I can't reconcile this with: Induced drag decreases with (square of) speed (for constant lift), because at higher speed there is more air to accelerate, so it only needs to be accelerated by less. From F=ma, sure the acceleration has decreased and the mass increased, but the force remains constant and since you've defined the reaction to this force as induced drag I can't see how that explains how induced drag reduces with airspeed.

@jumblie, the argument is not about force, but about power. Instead of $F = ma$, we express the force as $F = \dot m\Delta v_{air}$, where $\dot m$ is mass flow rate and $\Delta v_{air}$ the increase in (vertical) speed of the air. Since $\dot m\propto v$, $\Delta v\propto\frac{1}{v}$. Now in the reference frame of the free flow, this acceleration requires power $P = ½\dot m \Delta v_{air}^2$ and since this has one more $\Delta v_{air}$ factor, $P\propto\frac{1}{v}$. Since $P=Fv$, $F_{d}=\frac{P}{v}$ and this has another inverse $v$ factor, we get $F_{d}\propto\frac{1}{v^2}$.

@jumblie, if we look at the reference frame of the aircraft instead, we observe that the mass of air being affected at any given moment is actually constant and so is the acceleration, $a$. This is normal acceleration, so it changes the direction of the flow. We know that $a = \frac{v^2}{r} = \omega^2 r = v\omega$ ($v$ is the velocity and $\omega$ is angular velocity). $a$ is constant, so $\omega\propto\frac{1}{v}$. Now the change in angle $\Delta\alpha=\omega\frac{s}{v}$, where $s$ is approximately the chord length (also constant), so $\Delta\alpha\propto v^{-2}$ and $F_d=F\Delta\alpha$.

@jumblie, I ignored the trigonometry, approximating $\sin\alpha=\alpha$ and $\cos\alpha=1$, which is reasonable for small angles. I did so consistently in both solutions.

Induced drag is really a way aerodynamicists account for the local flow direction change (net downward) in the immediate proximity of a finite wing with sharp trailing edge as it moves thru a fluid at a positive angle of attack.  Induced drag is a term that is convenient as it implies a drag-due-to-lift origin and it does contribute to the total force buildup opposing motion.  But induced drag does not arise as a surface normal force nor skin traction force.  It is thus unlike form drag, skin friction' and excrescence drag which arise due to viscous shearing interaction between fluid and wing surface.  These drag forces, measurable in the wind tunnel, act on the wing skin as either a pressure-area normal force or tangential traction forces.  Form drag requires a robust internal structure to resist it;  skin friction peels paint;  and excrescence drag requires strong mounts for antennae and other protuberances. Induced drag is not measured in the wind tunnel;  it is calculated after the fact.  It and does not require structural accommodation to resist a force because it is not a force.
Of what physical nature, then, is it? It is actually the result of local distortion of the horizontal-vertical rectilinear reference framework that engineers have adopted for force accounting in airplane performance work.  By acclaim, the reference used to account for motion-retarding effects is parallel to the velocity vector at infinity, and all lifting effects are reconciled perpendicular to that vector.  Because of the downward flow component at the wing itself, the actual local velocity vector is inclined at a slight downward angle to the vector at infinity; the lift vector which is always normal to the actual local airflow, is thus angled backwards at the same angle.  When this 'bent-backwards' lift vector is reconciled into the force reference system (velocity st infinity), there will thus be a component of lift (its sine) aligned in the drag direction.  This we call induced drag.
It is important to understand the physics of this conveniently-named, but mis-informing term, to grasp what it means to airplane design and operation.
An English physicist-engineer-mathematician named Lanchester figured this out and published it in his textbook, 'Aerodonetics', before the Wright Brothers even flew at Kitty Hawk.

In short, no.  Induced drag is a byproduct of moving an airfoil through a fluid to create lift.  It is the sum of both the form drag of the airfoil as well as the unit vector of the lift line opposing the thrust vector of the propellor or jet exhaust.


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Does 14 CFR 25.1310 apply to hydraulic systems?
I'm struggling to understand if this particular rule was intended to apply to Electrical Power Generation Systems only or if this rule was also intended to apply to Hydraulic Power Generation Systems.
At first glance I assumed only electrical power generation systems would be applicable.  Then when I go to the EASA AMC for 25.1310 which reads the same it says the following:

"When alternative or multiplication of systems and equipment is
  provided to meet the requirements of CS 25.1310(a), the segregation
  between circuits should be such as to minimize the risk of a single
  occurrence causing multiple failures of circuits or power supplies of
  the system concerned.  For example, electrical cable bundles or groups
  of hydraulic pipes should be so segregated as to prevent damage to the
  main and alternative systems and power suppliers."

So EASA puts the words "hydraulic tubes" in as part of their example.
Is there some other guidance material out their that would clarify if the reg really applies to 25.1310?

The wording seems to be relevant to eliminate cause of common failures for all power generation systems.

Usually this link below can be very helpful in finding guidance for a particular regulation (e.g. 25.1310). In this case there isn't any guidance for 25.1310.
https://www.faa.gov/aircraft/air_cert/design_approvals/transport/rules_acs_policy/
Next try the preamble for when this regulation was passed or introduced and see if the FAA provided any guidance on their intent for the regulation. The FAA discussed an EWIS (Electrical Wiring Interconnect System) accident which only confuses the intent. However I briefly perused the preamble.
http://rgl.faa.gov/Regulatory_and_Guidance_Library/rgFinalRule.nsf/0/81B77634BED9EAA78625738E007123B3?OpenDocument&Highlight=amdt.%2025-123
25.1310 is in Part 25 Subpart F and most of these regulations (e.g. 25.1309) apply to all systems (e.g. hydraulic, electrical, pneumatic, etc.). Best of luck I hope this helps.


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Why is KOAK's alternate minimum higher than standard?
Why is KOAK's alternate minimum higher than standard?

According to TERPS (FAA Order 8260.3C) section 3-4, alternate minimum for PA is the higher of PA minimum or standard alternate minimum(600'-2).
KOAK ILS 12 minimum is 200'-1/2. Standard alternate minimum is higher than that PA minimum. Thus, alternate minimum should be 600'-2(standard). But actually, alternate minimum for KOAK ILS 12 is 1400'-3(for CAT D aircraft).

I'd like to know how alternate minimum to be determined.

Hi.  Welcome to Aviation.SE.  People volunteer and spend their own time helping others on this site.  It is normally expected that you will put at least as much effort into creating a good question as people will spend in answering it.  As it stands, it is unclear what you are asking and your question will probably be closed as such.  Please edit your question to include the  relevant detail and highlight exactly what you are not sure about.

I'm not sure how much detail you want, but see 3.4 in the FAA's TERPS. They calculate the minimums as usual (considering terrain, obstacles, weather etc.), and if the resulting numbers are higher than a set of standard reference numbers, alternate minimums are required. It's often difficult to find out more about one specific airport or approach because the calculations are complex.

It is unclear what you are asking by this question. I understand that you are wondering why KOAK's alternate minimums are high, presumable because you recognize that none of the airport's approaches are authorized for use for alternate planning. Why they are not authorized may be answerable. But what aspect of determining alternate minima are you asking about?

So let's discuss about KOAK's alternate minimums as a example. According to FAA Order 8260.3C  Section 3-4, alternate minimum for PA is the higher of PA minimum or standard alternate minimum(600'-2) .  For example, KOAK ILS 12 minimum is 200'-1/2. Standard alternate minimum is higher than that PA minimum. Thus, alternate minimum should be 600'-2(standard). But actually, alternate minimum for KOAK ILS 12 is 1400'-3(for CAT D aircraft). I'm very comfused.


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Why are RC rotor blades different from helicopter blades?
Why do toy (RC) helicopters have curved flat surfaces and larger rotors on true helicopters have a solid aerofoil shape?
If the rotor was mechanically strong enough to lift the desired weight, is there any (aerodynamic) advantage to the solid aerofoil over the simpler curved surface?
Surely toy manufacturers could "close the bottom" of the rotor (equating to the aerofoil shape) even if it was hollow with negligible increase in weight.
I understand the Bernoulli effect does not necessarily apply to the RC rotors and it may be simply a Coanda effect but they do seem to lift well with just electric motors.
I have seen discussion that there is a "scale up" effect but why and if so at what size does the simple rotor start to fail... one foot, three feet, ten feet? Is there any data to indicate when the transition from this simpler design to a fully extruded solid rotor is required?

Welcome to aviation.SE! I think I understand your question, but adding images (or links to some) would be very helpful

One factor that comes to mind is that in a full-sized helicopter the leading and trailing rotors have to have different angels of attack depending on the helicopters velocity. While that may be possible to accomplish on an RC helicopter, the cost would be prohibitive.

Thank you for your feedback. The first and second paragraphs are really the same question from different aspects. Why does the typical curved plastic blade of an RC helicopter not scale up to larger helicopters or conversely why don't RC helicopters not have aerofoil shaped rotors like larger helicopters of they are the most efficient shape. If the explanation is that there is some effect that "kicks in" as the scale increases, what might that be and when does it kick in?

Thank you I agree there would be an issue of differential lift if the RC helicopter moved forward quickly enough in a single rotor design but it is not the shape of the rotor that causes that. Many are coaxial to solve the problem just as with some full scale helicopters.

If the rotor was mechanically strong enough to lift the desired weight

It wouldn't be. Because of the square-cube law, lift and mechanical strength grow with second power of linear size, but weight grows with third power. The result is that at RC-scale, everything has plenty of power and strength even if made sloppily (and thus cheaply) from common materials while human-carrying-scale is really stretching capabilities of the materials and engines we have.
A highly cambered airfoil would not be strong enough for full-scale helicopter blade.

is there any (aerodynamic) advantage to the solid aerofoil over the simpler curved surface?

I believe there is. No modern aircraft uses highly cambered airfoil.

Surely toy manufacturers could "close the bottom" of the rotor (equating to the aerofoil shape) even if it was hollow with neglible increase in weight.

No. Efficiency is not that important at RC scale and hollow plastic is much more difficult to make (remember, plastic is made by injecting it into a mold).

I understand the Bernoulli effect does not necessarily apply to the RC rotors

Airfoils is an airfoil whether it is flying straight or rotates. And since Bernoulli's principle is just conservation of energy for fluid flow, it always applies to it. It does not, however, mean concave lower surface would mean higher flow speed and lower lift, because the length of the path has nothing to do with lift. The air above the wing is faster—and reaches the trailing edge long before the air below—for completely different reasons. Highly cambered wings do have higher lift. They just have even higher drag.

it may be simply a Coanda effect

Coandă effect is about a fluid jet surrounded with still fluid around it, but there are no jets of air involved here, so it can't be Coandă effect.

but they do seem to lift well with just electric motors

That's the square-cube law again making the aerodynamic performance so much better relative to weight at RC scale.

I have seen discussion that there is a "scale up" effect but why and if so at what size does the simple rotor start to fail

It is not just scale up (square-cube law):

Quad-copters with fixed pitch rotors (typical of the toy ones) are not controllable with failed engine. Human-carrying craft have to be.
Fixed pitch rotors deal poorly with horizontal speed, because the lift asymmetry creates big loads. Large rotors need to lead-lag and flap to compensate for this.

Both making quadcopters impractical at large scales.
Usually, quad-copters are made up to low tens of kilograms maximum weight. But where exactly is the limit might be hard to tell since larger ones may not be made simply because there is not much use for them.

It should be taken into account that the centrifugal force adds rigidity to the blades. In the early blimps, sometimes the propeller blades were made of canvas, and they assumed its 'working shape' only when spinning...

@xxavier, I don't think that would have effect on the cross-section of the blades (on helicopters) though.

No, but it's not necessary to build the blade of any highly resistant material, as it will work in tension only.

Thank you again. Do we know at what point we have to transition from the highly cambered aerofoil to the solid rotor?  Also, not trying to be contraversial, but when I was at NASA I saw some compelling data that the Bernoulli effect is in fact minor and it is actually the air "thrown down" by the Coanda effect that actually causes lift in almost all wing shapes. If so, then there would be no advantage to extruding a plastic wing over making a simple curved blade. I agree with the comments about fixed pitch but again that does not relate to rotor shape a fixed aerofoil would suffer the same.

Thanks. Yes rotating wings are more complex than the rules that guide the strength in other engineering areas. Parasails are flexible wings that use anchoring and air pressure. A spinning surface (like a garden trimmer thread) exposed to centrifugal force and surrounded by dynamic forces of air pressure above and below has greater mechanical strength than when it is static. But it is still hard to imagine you could make a rotor out of cloth but we see wings made out of cloth flying everyday and a rotor is a wing moving in a circle... interesting experiment. Maybe Da Vinci had it right.

Unfortunately, the paper I'd reference for this is behind a paywall, but there has been a significant experimental and computational study done showing that, under a Reynold's number of ~80,000 or so (which the rotors on some of these quadrotors fall into), a 3-6% thick, cambered flat plate is actually substantially MORE efficient (in terms of lift to drag ratio) than a traditional shapes.  However, I don't think this has been the actual design motivation.  The paper is here: https://vtol.org/store/product/basic-understanding-of-airfoil-characteristics-at-low-reynolds-numbers-11871.cfm

@CAerospace, lift is indeed because the air follows the wing, turning down in the process. However, this is not Coandă effect, only similar to it, because that talks about stream moving relative to the surrounding air and there is no separate stream here. There is also no separate contribution from “Bernoulli effect”. Bernoulli's principle, but is simply part of the one phenomenon. See also this answer.

@CAerospace, Both highly cambered thin airfoil and thicker airfoil produce lift, the difference is in how much drag they have. And as Marius notes, this depends on Reynolds number, so the highly cambered thin airfoil can be better at smaller scales, but not larger. There is no sharp limit; as the scale grows, both aerodynamic and structural constraints change continuously.

For doubters, this two photos show that cloth propellers existed, and worked. It's from a 1906 'Parseval' blimp... http://imgur.com/5IvtcpR

Ooops! Comment retracted.

This answer deserves +5 for how many questions it answered.

The two main differences are disk loading (lift per rotor disk area) and Reynolds number. The Reynolds number is the ratio between inertial and viscous forces.
Model helicopters operate at Reynolds numbers typical for small birds. If you look at their wings, they have a thin, cambered cross section. At this scale, loads are low and so they can afford to have thin wings. This reduces the acceleration of the flow from the displacement effect and helps to delay flow separation because the pressure gradients for the same lift are lower. At the low Reynolds number of small model helicopter rotors, the boundary layer will be fully laminar.
Full-scale helicopters, on the other hand, need much stronger rotor blades because of their higher disk loading and centrifugal loads. They cannot afford to use the thin airfoils of small birds, and the boundary layer can tolerate much higher pressure rises because the operating Reynolds number of a full-size helicopter rotor is large enough to ensure a turbulent transition.
Compared to wings, rotors enjoy a huge advantage: Flow separation is almost impossible. Centrifugal forces will push any slowed-down boundary layer out, where the higher circumferential speed will ensure that the flow at the wall never comes to the standstill required for flow separation.
Another reason for the flat shape of especially cheap indoor model helicopters with injection molded blades is the lower material use and shorter cycle times possible with a thin blade - they are much cheaper to produce.

Thank you. Some work done at Langley in the 80's showed that the Reynolds number is always above critical ( millions) in rotors but that scaled models could be used to predict full scale very accurately. The question I have is where do we transition from Hummingbird wing to Blackhawk rotor.

Indeed, even experiments with turbulator strips on rotor blades are rare.


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Do multi engines face the torque effect?
Do multi engines face the torque effect? What about jets?

By "torque effect" do you mean p-factor? Or some kind of rolling effect due to a high power engine?

@RonBeyer to be fair, if the OP understand the difference between torque and p-factor, he would not be asking this question.

I know that the torque effect is the left-turning tendency an aircraft has. Don't know what p-factor is (the Wikipedia explanation is a bit complicated). Any videos?

With all engines operating or one or more engines inoperative?

Maybe you mean the p-factor. If both engines are rotating clockwise, the plane yaws to the right. If one rotates clockwise, the other counter-clockwise, the effect cancelles out and the plane flys straight.

What about the effect of the two engines operating at different speeds? Is that significant?

@TannerSwett I think so. The p-factor is directly proportional with the speed of the prop. Now, if one engine operates at a different speed, the plane will yaw into the direction of the faster engine. (Apart from the fact that the thrust is applied asymmetrically)


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Why were canards added to some variants of the Dassault Mirage III?
Mirage III offshoots like Kfir made by Israel Aircraft Industries:

or Cheetah converted by South-African manufacturer Atlas Aviation:

both have canards.
The original Mirage III designed by French manufacturer Dassault has no canards:

What is the reason for such addition in variants?

Source for all pictures.

I guess a better question would be what was so compelling to add them on the derivative aircraft. Explaining a lack of something may be difficult: if the aircraft performs as designed to, why add more complexity? Engineers don't (usually) add complex stuff just for fun.

Probably for maneuverability.

The original delta Mirage was designed as an interceptor, basically to get there as soon as possible and try to stop a threat such as a nuclear bomber.  Speed was important above all else (except of course having enough stability to fly, where deltas also excel).
Canards are added for maneuverability for use as a point defender. As range is not critical for this task, the design remains viable today, but is being supplemented with stealth.
Many of the older reconnaissance/interceptors, such as the A-12/SR71 and the F15 Eagle, are still among the fastest ever flown.

and of course there was an idea to add winglets to the F-15 but it didn't make it past the state of a few (maybe even a single) technology demonstrator.

Also one could add that improved computer control (relaxed stability) allowed them to added as upgrades to older airframe designs.  Their effect on top speed would make an interesting study.

Because nose has heavy radar inside , the nose down at Mach tuck at supersonic is done by the canard

Do you have any sources to back up that as the reason they were added? I'm not saying it isn't, but let's have proof, please.


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Is there a Amelia Earhart anniversary late August 2024?
Today it came to my attention that in the last few days (late August 2024), there have been two concept albums centred on Amelia Earhart's last flight announced and shortly to be released by Laurie Anderson and Public Service Broadcasting
Neither mention why they released at this date.
Is there some anniversary of some event that would make sense of the timing or is in just a coincidence? Has some copyright expired on original recordings?

I think the speech on the (excellent) Public Service Broadcasting track is a recording artist that they used specifically for the track. It's not an original of Earhart's voice. So, the copyright idea might not be right. Source: interview with PSB on BBC Radio 6Music, in session with Steve Lamacq, 30th September (yesterday, as I write this!) https://www.bbc.co.uk/sounds/play/m00234hb?partner=uk.co.bbc&origin=share-mobile

(should have mentioned: the PSB track about Earhart that I'm referencing is called "Electre")

Although there are no officially stated holiday for Amelia Earhart in late August, there is an official holiday for her on July 24 every year, the day she was born.
There is one possible reason for the August releases, though. In August 1928, Earhart became the first woman to fly solo across the North American continent and back. Thus, it is possible that these albums decided to commemorate her by her achievements rather than birthday. This is just a guess, though, and the release dates may well have been completely unrelated to any holidays or commemoration days.


text:
What is high altitude testing?
When new engines are designed for aircraft (military or civilian) they undergo high altitude testing (especially the military ones).
What is high altitude testing of engines? How is that important?

High altitude means low air density. Ignition is harder to achieve with lower density, and the size of the combustor determines up to which altitude and throttle setting an engine will stay lit. Testing that it really works at those low densities before flying is a really good idea.

....and also explores the altitudes at which an engine re-light is possible.

Do they actually have to mount the engine to a test-bed plane and fly it at high altitude, or can they simulate the altitude in a test chamber on the ground? Or should I ask a new question?

@FreeMan: Both! Initially the test chamber is used, but later a multi-engined plane will carry one of the new ones aloft.


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How is supercruise achieved?
Wikipedia article tells that supercruise is a condition when an aircraft can achieve velocities above Mach one without using the highly inefficient afterburners. But still, there are not a lot of engines which can achieve this.
What is the specific requirement for an engine to achieve supercruise? Why can't most of the engines reach supercruise without reheats enabled?

It's less about the engine and more about the combination of engine and plane. Theoretically, all that is needed is for the engine(s)' thrust without afterburners to be able to overcome the plane's inherent drag at a speed greater than Mach 1. Simple, right?
Well, to do that, you're up against some pretty heavy physics. First and simplest, given a drag coefficient for the aircraft in a "clean" configuration, the drag force increases on the square of velocity. So, to go twice as fast, you need four times the thrust with all other things being equal. Jet engines simply don't scale that way; you need a bigger engine to provide more max thrust, and then all things aren't equal anymore.
Second, planes aren't built like cars, where in many cases there's enough room under the hood to drop in a bigger engine block and a turbocharger to double power for negligible additional weight. Fighter designs have to essentially wrap an aerodynamic skin as tightly as possible around the pilot, engine, avionics and fuel storage. For the longest time, engine efficiency was pretty static, therefore more thrust meant more engine. A bigger engine means a bigger plane, which means more drag, which offsets the thrust gains and thus the speed increase of the larger engine.
Thirdly, there are some complex aerodynamic changes involved as a plane transitions the sound barrier. Airflow into the engines increases at higher speeds, which reduces the compression ratio of the engine; the intake turbines become less and less efficient as speed increases, and thus you get less and less benefit from the "bypass" airflow of the turbofan. Afterburners are more or less a simple way to continue to get more thrust after the intake turbines are no longer significantly compressing air, by taking that air, dumping a bunch of extra fuel in it and riding the prolonged explosion. This, however, uses ten to fifty times the aircraft's cruising fuel flow rate.
The F-22 achieves supercruise in part by extending the efficient operating range of the engines with "variable bypass". At lower airspeeds, engine bypass is increased to allow the turbofans to increase thrust with minimum fuel. At higher airspeeds when the turbofan effect decreases, compressor bypass is reduced to put more air into the combustion chamber (along with more fuel), which increases thrust beyond the normal compression limit more efficiently than simply burning the bypass airflow with an afterburner.
Finally, there is the simple requirement of supercruise in an aircraft's mission profile. Not many things in air combat occur at supersonic speed; you can't dogfight effectively at Mach 1 because your sustainable turn rate is measured in miles, you often can't drop gravity ordinance (bombs) because they won't fall away from the aircraft in a stable or predictable way, and in the case of the Raptor you have to open payload doors to deploy your weapons and that will increase your drag right there.
Supercruise has a singular purpose; to increase the speed of the aircraft's transit to and from the target zone without sacrificing range by dumping fuel into the afterburners at ten to fifty times "military power". For an air superiority fighter (combination interceptor/dogfighter), that's very important, but the F-22 was the first U.S. fighter developed from the ground up to take advantage of the latest technological progress in aircraft design such that it could be achieved. The last time U.S. fighter designers got an opportunity to design a "cost-is-no-object" superfighter was almost 40 years ago with the F-15, and the two major fighter designs to enter U.S. service in the interim, the F-16 and F/A-18, were the result of a design competition specifically intended to create a lower-cost multirole fighter to serve as the bulk of the air fleet.
Other supercruise-capable fighters designed by NATO allies reached production first; the English Electric Lightning interceptor is considered the first supercruise-capable production fighter (predating the coining of that term), and the Dassault Rafale is reported to be supercruise-capable in certain mission-ready configurations up to Mach 1.3, and beat the Raptor to market by about 2 years, though the Raptor has a substantally higher top speed under full military power (100% non-afterburning throttle).

Actually, at higher speed the compression ratio increases, because the intake precompresses the air by decelerating it already. To the point that at high speeds a ramjet can do with just that and does not need any turbomachinery at all (see here). What reduces the effectiveness is that the exhaust speed does not grow as much as flight speed (if at all), so the change of speed decreases. Which is bigger problem for the bypass air that gets less speed to begin with.

We should not forget Concorde, which super cruised at Mach 2 for hours, and decades before the F-22.

actually one reason for supercruise is to be able to approach enemy fighters fast , to give ones own missiles maximum kinematic advantage, and to run from enemy missiles fast, to give them maximum disadvantage. Afterburner can be used to achieve this too, but the problem is that it eats fuel AND the ionized shock cones apparently increase  RCS to the point of being a problem.

Almost any jet engine can be used for super cruise. Certain characteristics help, however:

Convergent-divergent nozzle for supersonic exit velocities
Efficient intake with high pressure recovery
Good thrust/weight ratio

The aircraft needs to have sufficient thrust and low drag, so area-ruling and a low flight mass help.
Supercruise was invented as a marketing term for the Lockheed F-22. Note that a number of earlier designs could also cruise supersonically without afterburner. The stealth design of the F-22 requires the internal storage of weapons, which makes achieving super cruise easier. External stores full of weapons and external tanks would make super cruise practically impossible.


text:
How does the Piper Archer III electrical system work?
I can't seem to understand how the electrical system works:
What does the ammeter display?
What is said on my POH is that the ammeter displays the alternator output. My question is what output is that? In amps?
Does the battery power the DC bus?
It seems the battery takes AC from the alternator, convert it to direct current, and then feed the entire electrical bus system while still being charged. Is that correct?

I think you misunderstand how the alternator, battery, and electrical system are connected. An alternator puts out DC voltage into the electrical bus. It is used to provide electrical power to aircraft systems, as well as to recharge the battery from cranking and pre-start light and radio usage.
You can see this in the wiring diagram in the POH:

You can see the alternator about 2/3 of the way up the diagram, just to the left of center. The 'B' lead coming out of it goes directly into the ammeter and to the DC power bus (the big bar along the top).
The path back to the battery is a little more complicated, but there where it says "Battery Feeder", you can see it goes through a 70A fuse, down to where it connects to the battery contactor. So when the alternator switch is on and the battery switch is on (closing the battery contactor), the alternator can charge the battery. [In small Cessnas this is a split master switch, but I'm not sure what it looks like in an Archer.]
To answer your question specifically about what the ammeter shows, this will give you the amount of current the alternator is providing in amps. After startup you should see slightly more than usual as the battery recharges. You will probably see the number go up while you are transmitting on the radio. If you see this number get really large for no reason, there may be a fault in the electrical system somewhere. If you see this number go to ZERO, it's possible that the alternator has stopped working completely and your lights and radios are now running on battery for a limited amount of time!
I hope that is helpful. Note that nothing here should suggest a specific course of action -- this is just general electrical system information. Consult the emergency procedures or an experienced flight instructor for specific information about your airplane.

Technically the alternator does put out an AC voltage, it just uses a rectifier and regulator to convert it over to DC. All of this is internal to the alternator though.

So, uh, it puts out DC voltage then.

When you get through the rectifier and regulator, yes, I know it sounds a little strange but it is important to know how the alternator works internally, the "alternator" part of it is exactly that, it produces an alternating current.

Yes, all of us who design electrical systems know that, but I feel like this is extraneous and confusing in this context. On this schematic, the "alternator" puts out DC voltage.

Oh I see, so the alternator gives power to the electrical bus system as well as charging the battery correct? If I turned off the battery master switch off, the electrical system would still be working right? (If the engine is running, of course).

Yes, although the battery can serve to smooth out the voltage coming from the alternator, so it's not a great idea to run that way.

Can you explain what you mean by "smooth out the voltage"? Also, if you don't mind, I'm reading the POH and there's something called a overvoltage relay. I would assume this is where if there was an overvoltage of any particular item that is receiving too much voltage, the overvoltage relay would pop the circuit breaker, correct?

This might be a "crowbar" type circuit like you describe. Can you give a page reference? You might want to make this a separate question though.

The battery can absorb spikes in voltage to some degree during events known as a 'load dump'. This is a little bit down the rabbit hole of power system design though.

Ha, I see, and it's briefly spoken on 7-9 of the POH with the last sentence. I'm struggling to understand what overvoltage relay actually does...I think.

This is probably part of the ALTERNATOR CONTROL UNIT that disengages the alternator if there is an overvoltage problem. It's not possible to be more specific or authoritative than that from the information given.

I see, by the way, I see you said the ammeter tells us the current in amps of what the alternator is outputting, but to clarify, the ammeter is measure the alt. output to the bus system plus the charging of the battery right?

That's right. The battery is just connected to the main bus, so the battery charging is included in the alternator load.

Daniel gave you an excellent, detailed description of how the Piper ammeter (load meter) works.
Now I'm going to dumb it down to "What the heck does this mean in the cockpit?" and make all the other EEs on the site cringe.
Let's start with the assumption that the ammeters in small aircraft are going to read current, in amperes DC. (All electrical explanations aside the alternator is, for our purposes, a DC power source: We're only ever going to see its output after rectification).

In Piper aircraft the ammeter functions as a load meter: It shows how much current (in amps) the aircraft is demanding from the alternator for all purposes: running instruments/avionics, lights, and charging the battery.

Practically that means that the ammeter shows the total load the alternator is supplying (electrical accessories and battery charging) at any given time.

It will read zero any time the alternator is not providing power - running off the battery on the ground or with the alternator switch off.
It will initially show a high current load when the alternator is brought on-line, and this will reduce once the battery charge has been topped off from starting.
It should always show a load comparable to the running electronic equipment. For example when you turn the pitot heat on the load should go up by the amount of current that draws (around 10A) and stay there until you turn it off.

The only practical difference in the Warrior III and Archer III from previous generations of Pipers is that the meter is digital now, so instead of a needle with zero on the far left and the maximum alternator output on the far right you have a nice LCD display.

By contrast Cessna ammeters are arranged as a charge/discharge meter: They show how much current is flowing To or From the battery, and have their zero point in the center. Positive numbers represent charging the battery, and negative numbers represent discharging it.

Practically that means once the battery has been recharged after starting the plane the Cessna ammeter will show zero, or close to it, under normal circumstances.

It will show a discharge (negative numbers) when the alternator is not providing power - running off battery on the ground or with the alternator switch off.
It will initially show a charge (positive numbers) when the alternator is brought on-line and will drop to zero once the battery is topped off from starting.
It will show zero (no charge or discharge) when everything is operating normally in flight.
It will "twitch" when you add or remove load from the bus as the battery initially supplies the current demand until the alternator/regulator compensates for the new load. Using the same pitot heat example as for the Piper the Cessna load meter will briefly show a discharge when you turn pitot heat on, and may briefly show a charge when you turn it off, but will return to zero once the bus is stable with the new load.

Both systems have advantages and disadvantages, but most importantly the troubleshooting procedures are very different (in a Piper zero is bad and means your alternator is offline, in a Cessna zero is good and means everything is working) so it's important to know which system your aircraft has and how to interpret what it's telling you.

Well, can't the answer be as simple as: "The ammeter shows whether the battery is charging or being discharged." (it shows current (Ampere)). If the engine and alternator provide more power (current) than the equipment uses, the displayed value will be positive, and negative otherwise.


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Why does fuel vaporization in a carburetor cause a temperature decrease rather than a rise?
So I know the most "prominent" conditions of carburetor icing is anywhere from 20 to 70 °F with visible moisture. There is also a tendency to have icing in the carburetor in the venturi throat due to a drop in temperature due to fuel vaporization.
The problem I'm having is if fuel is vaporized, wouldn't that be increasing the temperature? I don't understand how the temperature would drop.

Good question! There's two parts to the answer.

The first part is the ideal gas law.
The formula of the ideal gas law is $PV = nRT$, where

P = Pressure
V = Volume
n = amount of substance
R = the ideal gas constant, which can be ignored for our purposes
T = Temperature

Since we're ignoring $R$, we're left with $PV = nT$.
Now, say that you send some air through the throat of a venturi. Pressure decreases greatly (that's the point), volume stays close enough that we can leave it alone, the amount of molecules remain unchanged, so to keep the equation in balance, the temperature drops.

The second part is the latent heat of vaporization
In order for a liquid to change state and become a gas, a certain amount of heat is required. Consider a firefighter spraying water on a fire. The liquid water is heated by the fire and evaporates into steam. The energy of the fire is decreased by the amount of energy that went into the steam and which is now floating away into the atmosphere.
Now, replace the fire with your carburetor, and replace the water with fuel. The heat energy in your carburetor is transferred to the fuel in order to cause it to vaporize. The fuel gets hotter (but who cares, it's about to get burned in the cylinders anyway) and the carburetor gets cooler.

I'm no mathematician, but I took the presumptive step of changing "N" to "n" in your simplified (without "R") equation. If that's incorrect, please roll it back.

The fuel does not get hotter when evaporating. This is exactly why cooling by evaporating works: you can remove a lot of energy without increasing the temperature. It's not a magic heat engine that makes hot fumes and cold remaining liquids, which is how your answer describes it now.

@Sanchises - I think we may be saying the same thing using different words. I agree that the fuel can absorb heat energy without an increase in temperature, if that energy is used to vaporize the fuel. If you're talking about something else, please feel free to edit my answer or add your own.

@SteveV.: No, you're not saying the same thing. The fuel doesn't get hotter, it gets colder. And it's the cold vaporized fuel which then cools the carburetor in turn. The reason is that vaporization is adiabatic; there's no significant heat exchange during vaporization.

Hi Steve, the second part is by far more important. Maybe you might want to add some expressions from this Wikipedia article and the concept of latent heat to improve the answer.

Sorry - disagree with the previous answer.
There's really just one effect. Molecules generally attract each other, until they come very close. If there's (virtually) no heat, then the molecules will settle at this close distance. You now have a solid.
Heat the solid a bit, and the molecules will start moving, and will no longer be at these closest distances, but they'll generally stay fairly close still - but they'll have enough energy to change neighbors.
Heat the liquid even further, and you'll see some of the molecules fly off entirely, and disappear in the gas phase. They still attract each other, so the speed of each individual molecule is constantly varying.
Now these change-over points are interesting. At these points, we see that with heat added, the average speed of molecules doesn't change. The extra energy goes to increasing the distances between molecules.
What happens in a carburetor is fairly simple. The fuel line is under normal pressure, and the fuel molecules are packed in a fluid. As the fuel enters the carburetor, it encounters a lower pressure which allows the molecules to disperse. As this dispersion increases the distance, the attraction between molecules slows down the escaping molecules - this is evaporative cooling.
Exactly how much cooling you have depends on the attraction as function of distance, which depends on the exact molecules. And aviation fuel is a mixture of many different chemicals, so that's rather complex.  But on average, it's predictable enough for engineering purposes. But yes, you might need to heat the carburetor, else you'll get droplets of fuel instead of a vapor.

Well, it looks like one of us is mistaken, because I agree with everything you've written up to, but not including As this dispersion increases the distance, the attraction between molecules slows down the escaping molecules - this is evaporative cooling. Both of our answers are lacking in references, so I'll try to find some evidence for mine. If you look for some evidence for yours maybe together we can meet at the truth.

See http://physics.stackexchange.com/questions/286013/is-there-a-melting-analogy-for-evaporative-cooling - this is fairly basic thermodynamics.

Physics are based on models. Models are like opinions: everyone has one.